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Kobotan [32]
1 year ago
12

Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.

The total average flow rate of the falls is 2.80×10³ m³/s and its average height is 52.0 m (Niagara Falls Live, 2017). Given that the density of water is 1.00×10³ kg/m³, calculate the average power output of Niagara Falls.
Average power output: _________.
Physics
1 answer:
Vinil7 [7]1 year ago
5 0

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

Q=-P_{tot}=-(P_{2}-P_{1})

The potential energy for any point can be calculated as:

P=m*g*h

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

Q=P_{1}=m*g*h

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW

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Aleksandr [31]

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

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2 years ago
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Sort the following forces as relevant or not relevant to this situation. The symbols are defined as follows: normal force = n⃗ ,
Irina18 [472]

Answer:

it would be least to graetest

Explanation:

10-84

7 0
2 years ago
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
ipn [44]

Answer:

ball clears the net

Explanation:

v_{o} = initial speed of launch of the ball = 20 ms^{-1}

\theta = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

v_{ox} = initial velocity = v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}

t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}       Eq-1

Consider the motion of the ball along the vertical direction

v_{oy} = initial velocity = v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}

t = time of travel

Y_{o} = Initial position of the ball at the time of launch = 2 m

Y = Final position of the ball at time "t"

a_{y} = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\

Since Y > 1 m

hence the ball clears the net

7 0
2 years ago
A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
DiKsa [7]

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

3 0
1 year ago
What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu
Dmitrij [34]

Answer:

-3.25\times 10^6 J

Explanation:

We are given that

Work done by the system=4.5\times 10^5 J

Heat transfer into the system=U_1=3.2\times 10^6 J

Heat transfer to the environment=U_2=6\times 10^6 J

We have to find the change in internal energy

By first law of thermodynamics

\Delta Q=\Delta U+w

\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J

Substitute the values then we get

-2.8\times 10^6=\Delta U+4.5\times 10^5

\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J

Hence, the change in internal energy =-3.25\times 10^6 J

7 0
2 years ago
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